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No, the closure of the image of $f$ in $Y$ is never the Stone-Čech compactification of $X$ unless $X$ is empty. In particular, consider the element $a\in Y$ which is $1$ on every coordinate. Note th …

Certainly not in general. For instance, let $F$ be the inclusion of the category of finite-dimensional vector spaces (over some fixed field) into the category of all vector spaces. If you want $\ma …

The empty category trivially satisfies this (there are no functors at all from a nonempty category to the empty category), but no other such category exists. Let $A$ be any category with a terminal o …

The answer is no. Let $C$ be a category such that the unique map from $C$ to the terminal category is a composition of $n$ adjoints. Then $C$ has an object $x_0$ such that every other object of $C$ …

If $f_\ast$ has a right adjoint, it must preserve colimits and hence be right-exact. Thus a necessary condition is that the higher derived functors vanish. In particular, when everything is affine an …

The answer is no, because the nerve functor turns an adjoint pair of functors between categories into inverse homotopy equivalences between spaces (this is because of the existence of the unit and cou …